Question

A car accelerates from rest with a 0.2m/^2 acceleration for 5 seconds and kept the same speed for 10 minutes and applied the brakes decelerate at 0.25m/s^2 till it came to a stop. How long is the trip?

Answer 1

My first question is are they asking for the length of time or the total distance?

Answer 2

total distance

Answer 3

I would use v = a*t to get the velocity that the car travels at for 10 mins
I would use s= 0.5*a*t^2 to get the distance traveled during the accleration

then the distance traveled for 10 mins: s= v*t

then find the time required to slow to 0 m/s: v= a*t
and then use t to find the distance traveled while decelerating.

Answer 4

oh ok that makes sense

Answer 5

one catch: velocity will be in m/s so you must change 10 min to 600 sec

Answer 6

ok thanks

Answer 7

to find the velocity I would use 0.2m/s^2 for a and 5 seconds for t?

Answer 8

yes,
\[ v= 0.2 m/s^2 * 5 s= 1.0 m/s \]

Answer 9

the second one would be 2.5 m for distance right?

Answer 10

yes, 2.5 is the distance traveled while accelerating up to 1 m/s
now stage 2, find the distance traveled during the next 600 sec

Answer 11

600m

Answer 12

yes

Answer 13

then what do I add to get the total distance?

Answer 14

you need to finish step 3
then find the time required to slow to 0 m/s: v= a*t
and then use t to find the distance traveled while decelerating.

Answer 15

for the time is it 5 or 4 seconds?

Answer 16

use decelerate at 0.25m/s^2 and v= 1 m/s in v= a*t

to be totally accurate, you are finding the change in velocity
\( v_f\) final velocity - \(v_i\) initial velocity= -0.25 m/s^2 * t
(it is -0.25 m/s^2 because you are slowing down)
of course \(v_f\) is 0 and \(v_i\) is 1 m/s

Answer 17

so its v= a*t
1= 0.25*t
4 s= t

then I add 600m + 4s =604 for the total distance

Answer 18

adding meters and seconds ??
remember, in step 1, you traveled 2.5 m while speeding up.

use the formula
d= 0.5*a*t^2
with a= 0.25 m/s^2 and t= 4 s
to find the distance traveled while slowing down

then add up the 3 distances.

Answer 19

oh ok

Answer 20

then is it 602.5?

Answer 21

that is the first 2 distances. what about the 3rd distance? (when slowing down)

Answer 22

its 603.5m

Answer 23

no, you must find the 3rd distance using
d= 0.5*a*t^2
with a= 0.25 m/s^2 and t= 4 s

Answer 24

the first distance = 2.5m
second= 600m
third= 2m

2.5 + 600 +2= 604.5

please say its right

Answer 25

yes, it is right

Answer 26

yes thank soo much
you helped me a lot